The magnitude of displacement vector of a particle which is moving in a circle of radius a with constant angular velocity ω as a function of time is

Question

The magnitude of displacement vector of a particle which is moving in a circle of radius a with constant angular velocity ω as a function of time is (A) 2 a text sin ω t (B) 2a textsin(ω t/2) (C) 2 a text cos ω t (D) 2acos(ω t/2)

Tardigrade Admin · Accepted Answer

In time t the particle has rotated through an angle θ =ω t <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-0uh9snliag3q.png" /> This means the magnitude of the displacement of the particle is s = textitPQ = √ textitQR2 + textitPR2 = √( textita sin ω t)2 + (a - texta cos ω t)2 s =2a textsin(ω t/2)

Q. The magnitude of displacement vector of a particle which is moving in a circle of radius $a$ with constant angular velocity $ω$ as a function of time is

$2 a sin ω t$

$2 a sin \frac{ω t}{2}$

$2 a cos ω t$

$2 a cos \frac{ω t}{2}$

In time $t$ the particle has rotated through an angle $θ = ω t$

This means the magnitude of the displacement of the particle is
$s = PQ = QR^{2} + PR^{2}$
$= (a sin ω t)^{2} + (a - a cos ω t)^{2}$
$s = 2 a sin \frac{ω t}{2}$

Q. The magnitude of displacement vector of a particle which is moving in a circle of radius a with constant angular velocity ω as a function of time is

2a sin ωt

2asin2ωt​

2a cos ωt

2acos2ωt​